# Submodule of a semisimple module is semisimple .

I have a doubt in proving the above statement . Lets say that $M$ a semisimple module , then consider $P \subset N$ , $P$ and $N$ are two sub-modules of $M$ .

now since $M$ is a semisimple module , then there exists $P'$ and $N'$ such that the direct sum of $P, P'$ is $M$ and $N, N'$ is $M$.

$N= M\cap N$ , now can i write $M$ as just $P+P'$ , i mean when can i write just sum in place of direct sum .

EDIT

My notes say $N= M\cap N =(P+P') \cap N = P+(P' \cap N)$ and $P\cap (P' \cap N) = \{0\}$ I know that $M$ is a direct sum of $P$ and $P'$ but my notes just igonres the direct sum sign . I don't understand whats going on .

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I cannot tell what you are asking, really. WHat role do $N$ and $N'$ play in all this? – Mariano Suárez-Alvarez Feb 7 '13 at 21:12
@MarianoSuárez-Alvarez : i will do some edit . – Theorem Feb 7 '13 at 21:15
Your edit did not help much... My mind reading maching suggests the following, though: if you know that $M=A\oplus B$, in particular you know that $M=A+B$, simply by the definition of what a direct sum is. – Mariano Suárez-Alvarez Feb 7 '13 at 21:27
@MarianoSuárez-Alvarez : that means the direct sum and sum are not different at all . – Theorem Feb 7 '13 at 21:30
No, it does not mean that. As I wrote, if you know that $M=A\oplus B$ then you know that $M=A+B$. I did not say the converse implication holds (and it doesn't) – Mariano Suárez-Alvarez Feb 7 '13 at 21:38

The simplest way to do this is based on the fact that

a module is semisimple if it is a sum of simple submodules.

If $M$ is semisimple, and $N\subseteq M$ is a submodule, then it is easy to check that $N$ is the sum of all the simple submodules of $M$ it contains, and therefore it is semisimple.

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There are a few equivalent ways to define a semisimple module and it would be helpful to know exactly which formulation you are using. One possible formulation is the following:

Definition. A module $M$ is semisimple if every submodule of $M$ is a direct summand of $M$.

Then to prove that every submodule of $M$ is semisimple let $N \subseteq M$ be a submodule and assume $P \subseteq N$ is a submodule of $N$. Write $M = N \oplus N' = P \oplus P'$ for some $N'$ and $P'$. What we need is to find $Q$ such that $N = P \oplus Q$.

Let $Q = P' \cap N$. This is a submodule of $N$ so we need to show that $N = P + Q$ and $P \cap Q = 0$. For the first let $n \in N$. Then $n \in M$ so $n = a + b$ for some uniquely determined $a \in P$ and $b \in P'$. As $P \subseteq N$ we have $n, a \in N$ therefore $b \in N$ therefore $b \in Q$. This gives $n \in P + Q$ and consequently $N = P + Q$.

For the second assume $n \in P \cap Q$. Then $n \in Q = P' \cap N$ so $n \in P'$. But $n \in P$ and $n \in P'$ imply $n = 0$. Hence $P \cap Q = 0$.

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Yes , this is more or less what i am following . My question is when can a person write just 'sum' instead of 'direct sum' ? – Theorem Feb 7 '13 at 21:35
When $A$ and $B$ are submodules of the same module $M$. – Jim Feb 7 '13 at 21:42
@Theorem Adding this because I'm speculating Theorem meant something else with that first comment. If $A$ and $B$ are submodules of $M$ and $A\oplus B=M$ is an (internal) direct sum, you can also write $A+B=M$. So, every (internal) direct sum equal to $M$ is also a sum equal to $M$, but the converse is not true. If $C+D=M$, the sum is not direct unless $C\cap D=\{0\}$. The following can be proven: a module is a sum of simple submodules iff it is a direct sum of simple submodules. – rschwieb Feb 7 '13 at 22:23